# Can any subset of $\mathbb{R}^2$ be expressed in form $A\times B$, where $A$ and $B$ are subsets of $\mathbb{R}$?

This is a very elementary question I'm a little confused about.

Can any subset of $\mathbb{R}^2$ be expressed in form $A\times B$, where $A$ and $B$ are subsets of $\mathbb{R}$?

I'm thinking that it might not necessarily be so. For instance, if we consider the intervals $[0,2]$ and $[3,5]$ in $\mathbb{R}$, and throw out a single point, say $(1,4)$, out of $[0,2]\times [3,5]$ then there do not exist subsets $A$ and $B$ of $\mathbb{R}$ such that $A\times B = ([0,2]\times [3,5]) \setminus \{(1,4)\}$. Because if they were equal, surely $A$ would be equal to $[0,2]$ and $B$ would be equal to $[3,5]$. But as there is no way to put the constraint that $1$ cannot pair up with $4$, $A\times B$ will always be $[0,2]\times [3,5]$.

Am I thinking in the right direction? Thanks.

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This is ideed not possible. – Michael Greinecker Feb 12 '13 at 17:08
Your argument works for me. Maybe the last sentence would rather be "so $A\times B=[0,2]\times [3,5]\neq [0,2]\times [3,5]\setminus \{(1,4)\}$, a contradiction". – 1015 Feb 12 '13 at 17:12
all $A\times B$ are rectangles. all subsets of $\mathbb{R}^2$ are not rectangles. – user59671 Feb 12 '13 at 17:14
@CutieKrait, this is not true. Take one of the sets not an interval, for example. – Sigur Feb 12 '13 at 17:15
@Sigur: finite or infinite unions of interiors of rectangles (with partial or complete borders or no border). – user59671 Feb 12 '13 at 17:39

If a subset $\mathrm X$ of $\mathbb R^2$ can be written as a product $\mathrm A \times \mathrm B$, then it is true that $\mathrm{pr_1}(\mathrm X) = \mathrm A$ and $\mathrm{pr_2}(\mathrm X) = \mathrm B$.

Then take the example of the unit disc in $\mathbb R^2$, it is not equal to $[-1,1]^2$ so it is not a product of two subset of $\mathbb R$.

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Are you considering the closed unit disc? If so, the projection on first coordinate will be onto the interval $[-1,1]$. The same for the second coordinate. The reason for the disc not be a product is other. – Sigur Feb 12 '13 at 17:26
Yes... What do you try to say ? – Damien L Feb 12 '13 at 17:29
If the set is a product $A\times B$ then its projections are $A,B$ but this could happen even if the set is not a product. So, why the disc is not a product? Note that its projections are intervals. – Sigur Feb 12 '13 at 17:32
@sigur that is precisely the point if his example - he provides a space that is clearly not $[-1,1]^2$, but which projects onto $[-1,1]$ and $[-1,1]$ in the two axes, so it is a counter-example. – Thomas Andrews Feb 12 '13 at 17:33
The point with coordinates $(1,1)$ has norm $\sqrt 2$ which is strictly greater than 1, so it is not in the unit disc. Therefore the two sets are not equal... – Damien L Feb 12 '13 at 17:39

Let $S:= \{ (x,y) \in \mathbf{R}^2 : x < y \}$, then I'm not sure if $S$ can ever be expressed as a Cartesian product of two subsets of $\mathbf{R}$.

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Take the unitary open disc, for example: $$D=\{(x,y)\in \mathbb{R}^2\mid x^2+y^2<1\}.$$

It is a good exercise to show that $D$ is not a product of any two subsets of $\mathbb{R}$. Here is a simple solution.

Suppose that $D=A\times B$, for some subsets $A,B\subseteq \mathbb{R}$. Note that $(x,0)\in D$ for any $x\in (-1,1)$. So $(-1,1)\subset A$ and $0\in B$. For the same reason, $0\in A$ and $(-1,1)\subset B$. Then $(-1,1)\times (-1,1)\subseteq A\times B=D$ and so take $\sqrt{2}/2< a,b < 1$ such that $(a,b)\in A\times B$ with $1< a^2+b^2$. Contradiction!

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For finite sets $A$ and $B$, the cardinality of $A\times B$ is the product of the two separate cardinalities. So for any set $S$ of $p$ points in the plane, $p$ being prime, the only way for $S$ to be a product set is for either the $x$-coordinates or the $y$-coordinates to be all the same.

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No. You can think of sets of the form $A \times B$ as 'rectangles' (although they are really only rectangles when $A,B$ are intervals). Not all sets are rectangles, for example, take the square and rotate it $45^\circ$.

To further illustrate, with the set $A \times B$, you can pick the element $a \in A$ and $b \in B$ 'independently' to get $(a,b) \in A \times B$, but this is not always the case. (For example, if I have the line $\{(x,y) | x = y\}$ and I choose $x=1$, then I must have $y=1$, whereas if I have the square $\{ (x,y) | x \in [0,1], y \in [0,1] \}$, then if I choose $x=1$, I can still pick any $y \in [0,1]$.)

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Yes you are thinking in the right direction. The answer is no. Here is a simple counter example: Consider the set $S = \{(1,2), (2,1) \}$. If there were sets $A,B \subseteq \mathbb R$ with $S = A \times B$ then $1,2 \in A$ but also $1,2 \in B$ so that $A \times B$ would have to contain $(1,1)$ and $(2,2)$ and therefore $A \times B$ is strictly greater than $S$.

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I like the fact that you thought of a finite set. There are infinite examples, but this is easy to see. – Ross Millikan Feb 13 '13 at 12:09